| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Prove/show always positive |
| Difficulty | Standard +0.8 Part (i) is routine substitution. Part (ii) requires solving a quartic that factors nicely. Part (iii) is the challenging element: students must form x^4 - kx^2 - 2 = 0, recognize it as a quadratic in x^2, and use discriminant analysis to prove intersection for all k—this requires insight beyond standard procedures and is non-trivial for C1 level. |
| Spec | 1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| \((0, -2)\) or 'crosses \(y\)-axis at \(-2\)' | B1 | |
| \((\pm 2^{\frac{1}{4}}, 0)\) | B2 | Or [when \(y = 0\)], \([x =] \pm 2^{\frac{1}{4}}\) or \(\pm\sqrt{\sqrt{2}}\) or \(\pm\sqrt[4]{2}\); B1 for one root correct |
| Answer | Marks | Guidance |
|---|---|---|
| \([y =]\ x^2 = x^4 - 2\) and rearrangement to \(x^4 - x^2 - 2 [= 0]\) or \(y^2 - y - 2 [= 0]\) | M1 | |
| \((x^2 - 2)(x^2 + 1) = 0\) oe in \(y\) | M1 | Or formula or completing the square; condone one error; condone replacement of \(x^2\) by another letter or by \(x\) for 2nd M1 (but not 3rd M1) |
| \(x^2 = 2\) [or \(-1\)] or \(y = 2\) or \(-1\) or \(x = \sqrt{2}\) or \(x = -\sqrt{2}\) | M1 | Dep on 2nd M1; allow inclusion of correct complex roots; M0 if any incorrect roots included for \(x^2\) or \(x\) |
| \((\sqrt{2}, 2)\) and \((-\sqrt{2}, 2)\); with no other intersections given | B2 | Or B1 for one of these two intersections (even if extra intersections given) or for \(x = \pm\sqrt{2}\) (and no other roots) or for \(y = 2\) (and no other roots) |
| Answer | Marks | Guidance |
|---|---|---|
| From \(x^4 - kx^2 - 2 [= 0]\): \(k^2 + 8 > 0\) | B1 | \(k^4 + 8k^2 > 0\); or from \(y^2 - k^2y - 2k^2 [= 0]\) |
| \(k + \sqrt{k^2 + 8} \geq 0\) for all \(k\) | B1 | \(k^2 + \sqrt{k^4 + 8k^2} > 0\) for all \(k\); [so there is a positive root for \(y\) and hence real root for \(x\) and so intersection]; if B0B0, allow SC1 for \(\frac{k \pm \sqrt{k^2+8}}{2}\) or \(\frac{k^2 \pm \sqrt{k^4 + 8k^2}}{2}\) obtained |
## Question 7:
### Part (i)
| $(0, -2)$ or 'crosses $y$-axis at $-2$' | B1 | | Condone $y = -2$ |
| $(\pm 2^{\frac{1}{4}}, 0)$ | B2 | Or [when $y = 0$], $[x =] \pm 2^{\frac{1}{4}}$ or $\pm\sqrt{\sqrt{2}}$ or $\pm\sqrt[4]{2}$; B1 for one root correct | |
### Part (ii)
| $[y =]\ x^2 = x^4 - 2$ and rearrangement to $x^4 - x^2 - 2 [= 0]$ or $y^2 - y - 2 [= 0]$ | M1 | | |
| $(x^2 - 2)(x^2 + 1) = 0$ oe in $y$ | M1 | Or formula or completing the square; condone one error; condone replacement of $x^2$ by another letter or by $x$ for 2nd M1 (but not 3rd M1) | If completing the square and haven't arranged to zero, can earn first M1 for an attempt such as $(x^2 - 0.5)^2 = 2.25$ |
| $x^2 = 2$ [or $-1$] or $y = 2$ or $-1$ or $x = \sqrt{2}$ or $x = -\sqrt{2}$ | M1 | Dep on 2nd M1; allow inclusion of correct complex roots; M0 if any incorrect roots included for $x^2$ or $x$ | NB for second and third M: M0 for $x^2 - 2 = 0$ or $x^2 = 2$ oe straight from quartic — some candidates probably thinking $x^4 - x^2$ simplifies to $x^2$; last two marks for roots are available as B marks |
| $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$; with no other intersections given | B2 | Or B1 for one of these two intersections (even if extra intersections given) or for $x = \pm\sqrt{2}$ (and no other roots) or for $y = 2$ (and no other roots) | Some candidates having several attempts at solving this equation — mark the best in this particular case |
### Part (iii)
| From $x^4 - kx^2 - 2 [= 0]$: $k^2 + 8 > 0$ | B1 | $k^4 + 8k^2 > 0$; or from $y^2 - k^2y - 2k^2 [= 0]$ | Condone lack of brackets in $(-k)^2$; alt methods: may use completing the square to show similarly, or comment that at $x = 0$ the quadratic is above the quartic and that as $x \to \infty$, $x^4 - 2 > kx^2$ for all $k$ |
| $k + \sqrt{k^2 + 8} \geq 0$ for all $k$ | B1 | $k^2 + \sqrt{k^4 + 8k^2} > 0$ for all $k$; [so there is a positive root for $y$ and hence real root for $x$ and so intersection]; if B0B0, allow SC1 for $\frac{k \pm \sqrt{k^2+8}}{2}$ or $\frac{k^2 \pm \sqrt{k^4 + 8k^2}}{2}$ obtained | |
---7
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d22f53f5-ba80-4065-a94b-2a9c92c20dfb-2_696_879_960_673}
\captionsetup{labelformat=empty}
\caption{Fig. 13}
\end{center}
\end{figure}
Fig. 13 shows the curve $y = x ^ { 4 } - 2$.\\
(i) Find the exact coordinates of the points of intersection of this curve with the axes.\\
(ii) Find the exact coordinates of the points of intersection of the curve $y = x ^ { 4 } - 2$ with the curve $y = x ^ { 2 }$.\\
(iii) Show that the curves $y = x ^ { 4 } - 2$ and $y = k x ^ { 2 }$ intersect for all values of $k$.
\hfill \mbox{\textit{OCR MEI C1 Q7 [10]}}